NERNST EQUATION AND CELL POTENTIAL CALCULATIONS

NERNST EQUATION AND CELL POTENTIAL

CALCULATIONS

1 INTRODUCTION

The Nernst equation relates the reduction potential of an electrochemical reaction to the

standard electrode potential, temperature, and activities (concentrations) of the chemical

species undergoing reduction and oxidation. It is fundamental in understanding the behavior of

galvanic cells and predicting the direction of spontaneous reactions.

2 THE NERNST EQUATION

The general form of the Nernst equation is:

𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

Where:

• 𝐸 is the cell potential under the specific conditions

• 𝐸∘ is the standard cell potential

• 𝑅 is the universal gas constant (8.314 J mol K−−1)

• 𝑇 is the absolute temperature in Kelvin

• 𝑛 is the number of moles of electrons transferred in the reaction

• 𝐹 is Faraday’s constant (96485 C mol−1)

• 𝑄 is the reaction quotient

3 PROBLEMS AND SOLUTIONS

3.1 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

1. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

2. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

3. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.2 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

1. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

2. 𝐸∘= 1.10 V (from Problem 1)

3. 𝑛 = 2 (two electrons are transferred)

4. 𝑇 = 25 °C+273.15 =298.15 K

5. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

6. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.3 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

1. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

2. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

3. 𝑛 = 2 (two protons transferred)

4. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

5. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.4 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

1. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

2. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

3. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

4. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

5. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.5 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

1. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

2. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

3. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

4. The negative value indicates that the reaction is spontaneous under standard conditions.

3.6 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

5. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

6. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

7. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.7 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

8. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

9. 𝐸∘= 1.10 V (from Problem 1)

10. 𝑛 = 2 (two electrons are transferred)

11. 𝑇 = 25 °C+273.15 =298.15 K

12. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

13. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.8 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

14. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

15. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

16. 𝑛 = 2 (two protons transferred)

17. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

18. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.9 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

19. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

20. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

21. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

22. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

23. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.10 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

24. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

25. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

26. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

27. The negative value indicates that the reaction is spontaneous under standard conditions.

3.11 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

28. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

29. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

30. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.12 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

31. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

32. 𝐸∘= 1.10 V (from Problem 1)

33. 𝑛 = 2 (two electrons are transferred)

34. 𝑇 = 25 °C+273.15 =298.15 K

35. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

36. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.13 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

37. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

38. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

39. 𝑛 = 2 (two protons transferred)

40. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

41. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.14 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

42. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

43. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

44. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

45. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

46. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.15 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

47. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

48. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

49. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

50. The negative value indicates that the reaction is spontaneous under standard conditions.

3.16 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

51. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

52. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

53. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.17 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

54. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

55. 𝐸∘= 1.10 V (from Problem 1)

56. 𝑛 = 2 (two electrons are transferred)

57. 𝑇 = 25 °C+273.15 =298.15 K

58. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

59. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.18 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

60. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

61. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

62. 𝑛 = 2 (two protons transferred)

63. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

64. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.19 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

65. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

66. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

67. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

68. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

69. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.20 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

70. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

71. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

72. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

73. The negative value indicates that the reaction is spontaneous under standard conditions.

3.21 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

74. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

75. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

76. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.22 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

77. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

78. 𝐸∘= 1.10 V (from Problem 1)

79. 𝑛 = 2 (two electrons are transferred)

80. 𝑇 = 25 °C+273.15 =298.15 K

81. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

82. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.23 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

83. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

84. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

85. 𝑛 = 2 (two protons transferred)

86. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

87. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.24 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

88. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

89. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

90. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

91. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

92. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.25 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

93. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

94. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

95. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

96. The negative value indicates that the reaction is spontaneous under standard conditions.

3.26 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

97. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

98. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

99. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.27 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

100. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

101. 𝐸∘= 1.10 V (from Problem 1)

102. 𝑛 = 2 (two electrons are transferred)

103. 𝑇 = 25 °C+273.15 =298.15 K

104. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

105. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.28 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

106. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

107. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

108. 𝑛 = 2 (two protons transferred)

109. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

110. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.29 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

111. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

112. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

113. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

114. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

115. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.30 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

116. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

117. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

118. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

119. The negative value indicates that the reaction is spontaneous under standard conditions.

3.31 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

120. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

121. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

122. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.32 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

123. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

124. 𝐸∘= 1.10 V (from Problem 1)

125. 𝑛 = 2 (two electrons are transferred)

126. 𝑇 = 25 °C+273.15 =298.15 K

127. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

128. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.33 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

129. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

130. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

131. 𝑛 = 2 (two protons transferred)

132. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

133. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.34 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

134. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

135. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

136. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

137. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

138. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.35 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

139. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

140. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

141. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

142. The negative value indicates that the reaction is spontaneous under standard conditions.

3.36 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

143. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

144. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

145. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.37 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

146. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

147. 𝐸∘= 1.10 V (from Problem 1)

148. 𝑛 = 2 (two electrons are transferred)

149. 𝑇 = 25 °C+273.15 =298.15 K

150. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

151. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.38 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

152. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

153. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

154. 𝑛 = 2 (two protons transferred)

155. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

156. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.39 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

157. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

158. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

159. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

160. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

161. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.40 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

162. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

163. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

164. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

165. The negative value indicates that the reaction is spontaneous under standard conditions.

3.41 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

166. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

167. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

168. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.42 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

169. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

170. 𝐸∘= 1.10 V (from Problem 1)

171. 𝑛 = 2 (two electrons are transferred)

172. 𝑇 = 25 °C+273.15 =298.15 K

173. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

174. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.43 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

175. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

176. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

177. 𝑛 = 2 (two protons transferred)

178. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

179. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.44 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

180. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

181. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

182. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

183. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

184. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.45 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

185. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

186. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

187. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

188. The negative value indicates that the reaction is spontaneous under standard conditions.

3.46 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

189. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

190. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

191. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.47 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

192. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

193. 𝐸∘= 1.10 V (from Problem 1)

194. 𝑛 = 2 (two electrons are transferred)

195. 𝑇 = 25 °C+273.15 =298.15 K

196. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

197. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.48 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

198. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

199. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

200. 𝑛 = 2 (two protons transferred)

201. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

202. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.49 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

203. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

204. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

205. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

206. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

207. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.50 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

208. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

209. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

210. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

211. The negative value indicates that the reaction is spontaneous under standard conditions.

3.51 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

212. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

213. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

214. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.52 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

215. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

216. 𝐸∘= 1.10 V (from Problem 1)

217. 𝑛 = 2 (two electrons are transferred)

218. 𝑇 = 25 °C+273.15 =298.15 K

219. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

220. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.53 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

221. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

222. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

223. 𝑛 = 2 (two protons transferred)

224. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

225. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.54 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

226. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

227. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

228. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

229. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

230. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.55 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

231. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

232. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

233. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

234. The negative value indicates that the reaction is spontaneous under standard conditions.

3.56 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

235. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

236. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

237. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.57 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

238. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

239. 𝐸∘= 1.10 V (from Problem 1)

240. 𝑛 = 2 (two electrons are transferred)

241. 𝑇 = 25 °C+273.15 =298.15 K

242. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

243. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.58 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

244. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

245. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

246. 𝑛 = 2 (two protons transferred)

247. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

248. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.59 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

249. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

250. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

251. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

252. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

253. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.60 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

254. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

255. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

256. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

257. The negative value indicates that the reaction is spontaneous under standard conditions.

3.61 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

258. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

259. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

260. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.62 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

261. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

262. 𝐸∘= 1.10 V (from Problem 1)

263. 𝑛 = 2 (two electrons are transferred)

264. 𝑇 = 25 °C+273.15 =298.15 K

265. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

266. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.63 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

267. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

268. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

269. 𝑛 = 2 (two protons transferred)

270. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

271. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.64 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

272. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

273. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

274. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

275. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

276. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.65 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

277. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

278. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

279. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

280. The negative value indicates that the reaction is spontaneous under standard conditions.

3.66 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

281. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

282. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

283. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.67 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

284. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

285. 𝐸∘= 1.10 V (from Problem 1)

286. 𝑛 = 2 (two electrons are transferred)

287. 𝑇 = 25 °C+273.15 =298.15 K

288. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

289. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.68 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

290. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

291. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

292. 𝑛 = 2 (two protons transferred)

293. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

294. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.69 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

295. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

296. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

297. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

298. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

299. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.70 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

300. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

301. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

302. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

303. The negative value indicates that the reaction is spontaneous under standard conditions.

3.71 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

304. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

305. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

306. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.72 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

307. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

308. 𝐸∘= 1.10 V (from Problem 1)

309. 𝑛 = 2 (two electrons are transferred)

310. 𝑇 = 25 °C+273.15 =298.15 K

311. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

312. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.73 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

313. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

314. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

315. 𝑛 = 2 (two protons transferred)

316. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

317. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.74 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

318. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

319. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

320. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

321. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

322. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.75 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

323. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

324. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

325. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

326. The negative value indicates that the reaction is spontaneous under standard conditions.

3.76 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

327. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

328. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

329. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.77 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

330. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

331. 𝐸∘= 1.10 V (from Problem 1)

332. 𝑛 = 2 (two electrons are transferred)

333. 𝑇 = 25 °C+273.15 =298.15 K

334. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

335. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.78 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

336. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

337. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

338. 𝑛 = 2 (two protons transferred)

339. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

340. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.79 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

341. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

342. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

343. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

344. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

345. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.80 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

346. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

347. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

348. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

349. The negative value indicates that the reaction is spontaneous under standard conditions.

3.81 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

350. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

351. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

352. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.82 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

353. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

354. 𝐸∘= 1.10 V (from Problem 1)

355. 𝑛 = 2 (two electrons are transferred)

356. 𝑇 = 25 °C+273.15 =298.15 K

357. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

358. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.83 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

359. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

360. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

361. 𝑛 = 2 (two protons transferred)

362. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

363. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.84 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

364. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

365. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

366. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

367. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

368. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.85 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

369. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

370. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

371. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

372. The negative value indicates that the reaction is spontaneous under standard conditions.

3.86 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

373. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

374. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

375. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.87 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

376. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

377. 𝐸∘= 1.10 V (from Problem 1)

378. 𝑛 = 2 (two electrons are transferred)

379. 𝑇 = 25 °C+273.15 =298.15 K

380. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

381. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.88 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

382. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

383. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

384. 𝑛 = 2 (two protons transferred)

385. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

386. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.89 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

387. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

388. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

389. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

390. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

391. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.90 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

392. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

393. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

394. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

395. The negative value indicates that the reaction is spontaneous under standard conditions.

3.91 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

396. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

397. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

398. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.92 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

399. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

400. 𝐸∘= 1.10 V (from Problem 1)

401. 𝑛 = 2 (two electrons are transferred)

402. 𝑇 = 25 °C+273.15 =298.15 K

403. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

404. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.93 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

405. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

406. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

407. 𝑛 = 2 (two protons transferred)

408. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

409. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.94 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

410. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

411. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

412. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

413. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

414. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.95 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

415. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

416. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

417. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

418. The negative value indicates that the reaction is spontaneous under standard conditions.

3.96 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

419. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

420. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

421. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.97 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

422. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

423. 𝐸∘= 1.10 V (from Problem 1)

424. 𝑛 = 2 (two electrons are transferred)

425. 𝑇 = 25 °C+273.15 =298.15 K

426. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

427. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.98 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

428. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

429. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

430. 𝑛 = 2 (two protons transferred)

431. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

432. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.99 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

433. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

434. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

435. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

436. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

437. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.100 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

438. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

439. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

440. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

441. The negative value indicates that the reaction is spontaneous under standard conditions.

3.101 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

442. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

443. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

444. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.102 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

445. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

446. 𝐸∘= 1.10 V (from Problem 1)

447. 𝑛 = 2 (two electrons are transferred)

448. 𝑇 = 25 °C+273.15 =298.15 K

449. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

450. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.103 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

451. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

452. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

453. 𝑛 = 2 (two protons transferred)

454. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

455. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.104 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

456. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

457. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

458. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

459. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

460. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.105 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

461. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

462. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

463. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

464. The negative value indicates that the reaction is spontaneous under standard conditions.

3.106 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

465. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

466. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

467. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.107 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

468. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

469. 𝐸∘= 1.10 V (from Problem 1)

470. 𝑛 = 2 (two electrons are transferred)

471. 𝑇 = 25 °C+273.15 =298.15 K

472. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

473. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.108 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

474. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

475. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

476. 𝑛 = 2 (two protons transferred)

477. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

478. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.109 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

479. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

480. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

481. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

482. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

483. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.110 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

484. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

485. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

486. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

487. The negative value indicates that the reaction is spontaneous under standard conditions.

3.111 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

488. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

489. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

490. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.112 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

491. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

492. 𝐸∘= 1.10 V (from Problem 1)

493. 𝑛 = 2 (two electrons are transferred)

494. 𝑇 = 25 °C+273.15 =298.15 K

495. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

496. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.113 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

497. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

498. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

499. 𝑛 = 2 (two protons transferred)

500. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

501. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.114 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

502. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

503. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

504. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

505. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

506. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.115 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

507. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

508. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

509. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

510. The negative value indicates that the reaction is spontaneous under standard conditions.

3.116 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

511. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

512. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

513. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.117 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

514. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

515. 𝐸∘= 1.10 V (from Problem 1)

516. 𝑛 = 2 (two electrons are transferred)

517. 𝑇 = 25 °C+273.15 =298.15 K

518. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

519. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.118 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

520. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

521. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

522. 𝑛 = 2 (two protons transferred)

523. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

524. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.119 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

525. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

526. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

527. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

528. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

529. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.120 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

530. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

531. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

532. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

533. The negative value indicates that the reaction is spontaneous under standard conditions.

3.121 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

534. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

535. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

536. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.122 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

537. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

538. 𝐸∘= 1.10 V (from Problem 1)

539. 𝑛 = 2 (two electrons are transferred)

540. 𝑇 = 25 °C+273.15 =298.15 K

541. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

542. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.123 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

543. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

544. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

545. 𝑛 = 2 (two protons transferred)

546. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

547. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.124 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

548. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

549. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

550. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

551. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

552. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.125 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

553. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

554. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

555. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

556. The negative value indicates that the reaction is spontaneous under standard conditions.

3.126 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

557. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

558. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

559. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.127 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

560. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

561. 𝐸∘= 1.10 V (from Problem 1)

562. 𝑛 = 2 (two electrons are transferred)

563. 𝑇 = 25 °C+273.15 =298.15 K

564. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

565. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.128 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

566. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

567. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

568. 𝑛 = 2 (two protons transferred)

569. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

570. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.129 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

571. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

572. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

573. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

574. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

575. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.130 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

576. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

577. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

578. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

579. The negative value indicates that the reaction is spontaneous under standard conditions.

3.131 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

580. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

581. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

582. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.132 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

583. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

584. 𝐸∘= 1.10 V (from Problem 1)

585. 𝑛 = 2 (two electrons are transferred)

586. 𝑇 = 25 °C+273.15 =298.15 K

587. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

588. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.133 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

589. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

590. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

591. 𝑛 = 2 (two protons transferred)

592. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

593. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.134 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

594. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

595. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

596. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

597. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

598. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.135 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

599. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

600. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

601. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

602. The negative value indicates that the reaction is spontaneous under standard conditions.

3.136 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

603. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

604. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

605. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.137 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

606. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

607. 𝐸∘= 1.10 V (from Problem 1)

608. 𝑛 = 2 (two electrons are transferred)

609. 𝑇 = 25 °C+273.15 =298.15 K

610. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

611. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.138 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

612. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

613. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

614. 𝑛 = 2 (two protons transferred)

615. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

616. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.139 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

617. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

618. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

619. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

620. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

621. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.140 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

622. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

623. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

624. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

625. The negative value indicates that the reaction is spontaneous under standard conditions.

3.141 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

626. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

627. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

628. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.142 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

629. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

630. 𝐸∘= 1.10 V (from Problem 1)

631. 𝑛 = 2 (two electrons are transferred)

632. 𝑇 = 25 °C+273.15 =298.15 K

633. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

634. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.143 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

635. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

636. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

637. 𝑛 = 2 (two protons transferred)

638. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

639. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.144 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

640. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

641. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

642. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

643. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

644. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.145 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

645. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

646. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

647. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

648. The negative value indicates that the reaction is spontaneous under standard conditions.

3.146 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

649. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

650. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

651. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.147 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

652. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

653. 𝐸∘= 1.10 V (from Problem 1)

654. 𝑛 = 2 (two electrons are transferred)

655. 𝑇 = 25 °C+273.15 =298.15 K

656. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

657. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.148 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

658. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

659. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

660. 𝑛 = 2 (two protons transferred)

661. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

662. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.149 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

663. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

664. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

665. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

666. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

667. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.150 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

668. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

669. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

670. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

671. The negative value indicates that the reaction is spontaneous under standard conditions.

3.151 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

672. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

673. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

674. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.152 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

675. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

676. 𝐸∘= 1.10 V (from Problem 1)

677. 𝑛 = 2 (two electrons are transferred)

678. 𝑇 = 25 °C+273.15 =298.15 K

679. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

680. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.153 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

681. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

682. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

683. 𝑛 = 2 (two protons transferred)

684. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

685. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.154 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

686. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

687. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

688. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

689. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

690. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.155 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

691. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

692. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

693. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

694. The negative value indicates that the reaction is spontaneous under standard conditions.

3.156 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

695. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

696. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

697. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.157 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

698. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

699. 𝐸∘= 1.10 V (from Problem 1)

700. 𝑛 = 2 (two electrons are transferred)

701. 𝑇 = 25 °C+273.15 =298.15 K

702. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

703. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.158 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

704. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

705. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

706. 𝑛 = 2 (two protons transferred)

707. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

708. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.159 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

709. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

710. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

711. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

712. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

713. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.160 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

714. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

715. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

716. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

717. The negative value indicates that the reaction is spontaneous under standard conditions.

3.161 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

718. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

719. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

720. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.162 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

721. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

722. 𝐸∘= 1.10 V (from Problem 1)

723. 𝑛 = 2 (two electrons are transferred)

724. 𝑇 = 25 °C+273.15 =298.15 K

725. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

726. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.163 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

727. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

728. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

729. 𝑛 = 2 (two protons transferred)

730. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

731. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.164 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

732. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

733. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

734. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

735. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

736. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.165 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

737. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

738. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

739. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

740. The negative value indicates that the reaction is spontaneous under standard conditions.

3.166 PROBLEM 1: STANDARD CELL POTENTIAL

Calculate the standard cell potential for the following reaction:

Given: 𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )=\𝑆𝐼−0.76\𝑣𝑜𝑙𝑡,𝐸 �𝑖𝑟𝑐(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )=\𝑆𝐼0.34\𝑣𝑜𝑙𝑡olution:

741. The standard cell potential is the difference between the reduction potentials of the

cathode and anode:

𝐸cell

∘= 𝐸cathode

∘−𝐸anode

∘

742. In this case, Cu(2+) is being reduced (cathode) and Zn is being oxidized (anode):

𝐸(\𝑡𝑒𝑥𝑡𝑐𝑒𝑙𝑙)

∘= 𝐸(\𝑐𝑒𝐶𝑢2+

𝐶𝑢 )

∘−𝐸(\𝑐𝑒𝑍𝑛2+

𝑍𝑛 )

∘

743. Substituting the values:

𝐸cell

∘= 0.34 V−(−0.76 V)= 1.10 V

3.167 PROBLEM 2: NON-STANDARD CONDITIONS

Calculate the cell potential at 25 °C for the reaction in Problem 1 when[\𝑐𝑒𝐶𝑢2+]=

\𝑆𝐼0.1\𝑚𝑜𝑙𝑎𝑟𝑎𝑛𝑑[\𝑐𝑒𝑍𝑛2+]=\𝑆𝐼0.01\𝑚𝑜𝑙𝑎𝑟.

Solution:

744. Use the Nernst equation: 𝐸 = 𝐸∘−𝑅𝑇/𝑛𝐹ln𝑄

745. 𝐸∘= 1.10 V (from Problem 1)

746. 𝑛 = 2 (two electrons are transferred)

747. 𝑇 = 25 °C+273.15 =298.15 K

748. The reaction quotient $Q = \𝑓𝑟𝑎𝑐{[\𝑐𝑒{𝑍𝑛^2+}]}{[\𝑐𝑒{𝐶𝑢^2+}]} = \𝑓𝑟𝑎𝑐{0.01}{0.1} =

0.1$

749. Substituting into the Nernst equation:

■(𝐸& = 1.10−(8.314)(298.15)/(2)(96485)𝑙𝑛(0.1)@& = 1.10+0.0296@& = 1.13 V)

3.168 PROBLEM 3: CONCENTRATION CELL

Calculate the cell potential at 25 °C for a concentration cell with two hydrogen electrodes,

where $[\𝑐𝑒{𝐻+}]_1 = \𝑆𝐼{1𝑒 − 3}{\𝑚𝑜𝑙𝑎𝑟}$ 𝑎𝑛𝑑 $[\𝑐𝑒{𝐻+}]_2 = \𝑆𝐼{1𝑒−5}{\𝑚𝑜𝑙𝑎𝑟}$.

Solution:

750. For a concentration cell, 𝐸∘= 0 since both half-cells use the same electrode

751. The cell reaction is: $\𝑐𝑒{2𝐻 + (\𝑡𝑒𝑥𝑡{𝑑𝑖𝑙𝑢𝑡𝑒}) −> 2𝐻 + (\𝑡𝑒𝑥𝑡{𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑})}$

752. 𝑛 = 2 (two protons transferred)

753. 𝑄 = [\𝑐𝑒𝐻+]1

[\𝑐𝑒𝐻+]2=10−3

10−5 =100

754. Using the Nernst equation:

■(𝐸& = 0−(8.314)(298.15)/(2)(96485)𝑙𝑛(100)@& = 0.0592𝑙𝑜𝑔(100)@& = 0.118 V)

3.169 PROBLEM 4: PH MEASUREMENT

A hydrogen electrode is coupled with a standard calomel electrode (SCE, 𝐸∘= 0.2415 V) to

measure pH. The cell potential is found to be 0.295 V at 25 °C. Calculate the pH of the solution.

Solution:

755. The cell reaction is: \𝑐𝑒2𝐻++2𝑒−→ 𝐻2

756. 𝐸cell = 𝐸\𝑐𝑒𝑆𝐶𝐸 −𝐸\𝑐𝑒𝐻+/𝐻2

757. Using the Nernst equation for the hydrogen electrode:

𝐸\𝑐𝑒𝐻+/𝐻2 = 0−0.0592log(1

[\𝑐𝑒𝐻+]) = 0.0592pH

758. Substituting into the cell potential equation:

0.295 = 0.2415−0.0592pH

759. Solving for pH:

■(0.0592pH& = 0.2415−0.295 = −0.0535@pH& = (−0.0535)/0.0592 = 7.04)

3.170 PROBLEM 5: GIBBS FREE ENERGY

Calculate the standard Gibbs free energy change for the reaction in Problem 1 at 25 °C.

Solution:

760. The relationship between Gibbs free energy and cell potential is:

𝛥𝐺∘= −𝑛𝐹𝐸∘

761. From Problem 1, 𝐸∘= 1.10 V and 𝑛 = 2

762. Substituting:

■(𝛥𝐺∘& = −(2)(96485)(1.10 V)@& = −212267 J/mol@& = −212.3 kJ/mol)

763. The negative value indicates that the reaction is spontaneous under standard conditions.

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